一个朋友最近让我编写一个算法,它取n个“人”,生成一个n-1乘n/2的网格,其中每个可能的对出现一次,在每个n/2部分中,任何人都不允许出现两次。(即person1与person2匹配,person1与person3匹配无效)。你知道吗
如果这不合理,想象一下这样的情景:为100个人创建99轮会议,每个人在每轮会议中都会遇到一个新的人,但没有人在一轮会议中有超过一个会议。这样,每轮举行50次独特的会议,共99轮,共计4950次独特的会议。你知道吗
我在python中使用递归回溯(不确定这是否是最好的方法)编写了这个代码。这个问题让我想起了数独,这是一个流行的数独解决方法)这需要一个愚蠢的时间,即使是很小的数字(例如50)。有没有一种更快的编码方法,或者我可以添加一些增强来让它运行得更快?你知道吗
代码在这里:https://pastebin.com/iyr2wdkz
import random
from itertools import combinations
import time as t
import sys
sys.setrecursionlimit(15000)
def isPrime(num):
"""Checks num for primality. Returns bool."""
if num == 2:
return True
elif num < 2 or not num % 2: # even numbers > 2 not prime
return False
# factor can be no larger than the square root of num
for i in range(3, int(num ** .5 + 1), 2):
if not num % i: return False
return True
def generatePrimes(n):
"""Returns a list of prime numbers with length n"""
primes = [2,]
noOfPrimes = 1 # cache length of primes for speed
testNum = 3 # number to test for primality
while noOfPrimes < n:
if isPrime(testNum):
primes.append(testNum)
noOfPrimes += 1
testNum += 2
return primes
class Person:
def __init__(self, name, ID):
self.name = name
self.primeID = ID
def __eq__(self, other):
return self.primeID == other
def __repr__(self):
return '%d' % (self.name)
class Schedule:
def __init__(self, numberofpeople):
self.x = 0 #current x pos
self.y = 0 #current y pos
self.fill = 0 #number of slots filled
self.board = [] #get initialized to an n-1 x n/2 grid to hold links
self.failed = [] #same as board but holds a list of failed links in each cell
self.uniqueLinks = [] #list of unique links, a link is the product of two primes
self.availableLinks = [] #links not yet placed in the grid
self.personList = [] #list of Person. A person is a name and a unique prime number
self.primeList = generatePrimes(numberofpeople) #list of the first n prime numbers
random.shuffle(self.primeList)
#initializes the empty lists
for i in range(numberofpeople):
self.personList.append(Person(i, self.primeList[i]))
self.uniqueLinks = list(map(lambda x: x[0] * x[1], combinations(self.primeList, 2)))
for i in self.uniqueLinks:
self.availableLinks.append(i)
tmp = len(self.uniqueLinks)
for i in range(tmp // (numberofpeople // 2)):
self.board.append(list())
self.failed.append(list())
for i in range(len(self.board)):
for j in range(numberofpeople//2):
self.board[i].append(None)
self.failed[i].append(list())
#checks if the candidate is valid in current position
def isValid(self, candidate):
factor1, factor2 = self.getFactor(candidate)
for i in self.board[self.x]:
if not i:
return True
if ((i % factor1) == 0) or ((i % factor2) == 0):
return False
return True
#moves to the next non-None value, return True if successful
def nextpos(self):
for i in range(len(self.board)):
for j in range(len(self.board[0])):
if not self.board[i][j]:
self.x = i
self.y = j
return True
return False
#sets the last non-None value to None and adds that value to availableLinks and the failed tracker
def prevpos(self):
for i in range(len(self.board)-1, -1, -1):
for j in range(len(self.board[0])-1, -1, -1):
if self.board[i][j]:
self.x = i
self.y = j
tmp = self.board[self.x][self.y]
self.availableLinks.append(tmp)
self.board[i][j] = None
self.failed[i][j].append(tmp)
self.fill -= 1
random.shuffle(self.availableLinks)
return True
#returns the prime factors of num
def getFactor(self, num):
for i in self.primeList:
if num % i == 0:
return i, num/i
#recursive backtracking solving algorithm
def solve(self):
print('%d links placed, %d remaining, pos is %d, %d' % (self.fill, len(self.availableLinks), self.x, self.y))
if not self.nextpos():
return True
for i in self.availableLinks:
if self.isValid(i): and self.checkFail(i):
self.fill += 1
self.board[self.x][self.y] = i
self.availableLinks.remove(i)
if self.solve():
return True
else:
self.prevpos()
return False
#replaces prime products with formatted names of people in the board
def decompose(self):
for i in range(len(self.board)):
for j in range(len(self.board[0])):
num1, num2 = self.getFactor(self.board[i][j])
for k in self.personList:
if k == num1:
person1 = str(k)
if k == num2:
person2 = str(k)
self.board[i][j] = person1 + '<-->' + person2
# checks if num has already been tried at current pos, returns false if is has.
def checkFail(self, num):
if num in self.failed[self.x][self.y]:
return False
else:
return True
# verifies that the board has no duplicate values
def verify(self):
visited = []
for i in self.board:
for j in i:
if j in visited:
print('Verification failed %s occurs twice'% j)
return False
visited.append(j)
print('Successfully verified')
return True
s =Schedule(50)
s.solve()
s.verify()
s.decompose()
print(s.board)
看来你需要round-robin tournament algorithm。你知道吗
简言之:创建两行表,固定第一行的第一个条目,并循环移动所有其他条目(n-1次)。在每一个圆对是由相同的列中的顶部和底部条目组成的
多么惊人的代码量!你知道吗
这是scala中的一个解决方案。我只使用int,你可以映射到它个人ID地址:
对于100人,它会立即返回(<;200毫秒)。你知道吗
为了适应更多的人,我会把它转换成一个尾部递归函数。你知道吗
想法是,第一个人向所有人打招呼,然后离开房间。然后第二个人向大家打招呼,也离开了。最后99人向100人打招呼,两人都离开了房间。你知道吗
然后他们开始吃喝。你知道吗
以下是一种非递归方法:
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